Answer:
b. Frank received the mutant chromosome from his father. Nondisjunction occurred in his father during the first meiotic division.
Explanation:
As you can see in the question above, Frank has Klinefelter syndrome which causes him to have normal skin patches and skin patches without sweat glands. Her mother has completely normal hair, which may indicate that the defective gene was not supplied by her. In addition, Frank's father has anhydrotic ectodermal dysplasia, an X-linked condition where the skin does not contain sweat glands.
Although Frank's father's defective gene is linked to the X chromosome, it is likely that Frank inherited the defective gene from his country. This may have occurred because during meiosis I, his father's genes did not show disjunction. As a result, Frank presents a mosaic of his phenotype, because an inactivation of the X chromosome occurred.
An organism described as 2n=4 has the chromosomes below with genes indicated by letters and centromeres indicated by periods. Select the BEST description of the chromosome aberration present in this organism.
HJM.NPQ HJM.NPQ R.STWXSTZ R.STWXZ
Select one:
a. Reverse Tandem Duplication
b. Terminal Deletion
c. Tandem Duplication
d. Heterobrachial Displaced Duplication
e. Reciprocal Translocation
f. Homobrachial Displaced Duplication
g. Nonreciprocal Translocation
h. Robertsonian Translocation
i. Interstitial Deletion
j. Paracentric Inversion
k. Pericentric Inversion
Answer:
k. Pericentric InversionExplanation:
Such type of the condition in which any abnormality takes place in the chromosomes due to deletion or addition of the nucleotide in the DNA of the chromosomes is known as a chromosomal aberration or chromosomal abnormality. Generally, it takes place due to, mutation process and is classified into different groups such as deletion, insertion, pericentric inversion, and some others.
Inversion is a type of chromosomal abnormality in which the genetic material is inverted and thus caused abnormality. When the inversion process occurs around the centromeres then such type of chromosomal abnormality is called a pericentric inversion.
The chromosomal aberration present in the organism is a Pericentric Inversion. This is a mutation where a segment of a chromosome including the centromere has reversed its orientation.
Explanation:This question pertains to identifying a chromosomal aberration, specifically by examining the genetic sequence of an organism that is described as having a 2n = 4 chromosome count. Observing the provided sequences, we can notice that in the third and fourth sequences of the chromosomes, there's a segment 'STWX' followed by another 'STXZ'. This repetition represents a kind of chromosomal mutation. This particular type of mutation is called a Pericentric Inversion.
A Pericentric Inversion is a chromosomal mutation or aberration where a segment of a chromosome which includes the centromere has reversed its orientation - hence the term inversion.
Learn more about chromosome aberration here:https://brainly.com/question/34764823
#SPJ3
Assume that you have an illuminated suspension of Chlorella cells carrying out photosynthesis in the presence of 0.1% carbon dioxide and 20% oxygen.
What will be the short-term effects of the following changes in conditions on the levels of 3-phosphoglycerate (PGA) and ribulose-1,5-bisphosphate (RuBP)?
Carbon dioxide concentration is suddenly reduced 1000-fold.
a. PGA down, RuBP up
b. PGA up, RuBP down
c. PGA up, RuBP unchanged
d. PGA unchanged, RuBP up
Answer:
a. PGA down, RuBP up
Explanation:
RuBP, also known as Ribulose-1,5-bisphosphate is a molecule that assist in he fixation of carbon dioxide in the Calvin Cycle or light independent reaction of photosynthesis.
RuBP is combined with carbon dioxide using the enzyme known as rubisco to form an short-lived intermediate product which divides to form two 3-carbon molecule structure known as 3-phosphoglycerate.
If the concentration of carbon dioxide is suddenly reduced, it thus means that the rate of production of 3-phosphoglycerate will reduce as carbon dioxide becomes a limiting factor. Hence, 3-phosphoglycerate's concentration will reduce while more RuBP will become available as a result.
The correct option is a.
Another name for an oxidizing agent is a(n) Group of answer choices electron acceptor. hydride transfer reagent. electron donor. electropositive metal. reductant.
Explanation:
an oxidizing agent is an agent that can oxidize other substances, therefore, accepting their electrons. common oxidizing agents are oxygen and hydrogen peroxide...
Answer: Option A.
Hybride transfer.
Explanation:
An oxidizing agent is a compound that accept electron and transfer oxygen atom during a Redbox reaction . An oxidizing agents oxidize others in a chemical reaction. Oxidizing agents are reduced in a chemical reaction. Oxidizing agents are electronegative substance. Oxidizing agents are hybrid transfer, they transfer oxygen atoms to other substances. Examples of oxidizing agents are; chlorine, flourine,oxygen e.t.c.
Along with a balanced diet, weight-bearing exercise promotes bone mineral density. Ruben's body-builder diet would be unhealthy for his bones over the long term because it is __________.
Answer: it is lacking adequate levels of Vitamin D and Calcium.
Explanation:Vitamin D deficiency can lead to loss of bone density, which is a contributory factor to osteoporosis and broken bones.
Severe vitamin D deficiency can also lead to other diseases. Vitamin D deficiency leads to osteomalacia. Osteomalacia which causes weak bones, bone pain, and muscle weakness. Because Reuben's body builder diet lacks Vitamin D, it can cause fractures and even osteoporosis over when used for a long time especially without trying to make amends for the deficient Vitamin D.
Answer:inadequate in vitamin D and excessive in calcium
Explanation:
Vitamin D plays a significant role in the regulation of calcium and maintenance of phosphorus levels in the blood. These factors are vital for maintaining healthy bones. People need vitamin D to allow the intestines to stimulate and absorb calcium and reclaim calcium that the kidneys would otherwise excrete.
Foods that provide vitamin D include:fatty fish, like tuna, mackerel, and salmon.
Foods fortified with vitamin D, like some dairy products, orange juice, soy milk, and cereals.
Beef liver.
Cheese.
Egg yolks.
17. Which of the following happens first as a nephron processes blood?
A. reabsorption
B. excretion
C. osmosis
D. filtration
E. secretion
Answer: Option D) Filtration
Explanation:
Blood brought to the kidney by renal artery. As it circulates through the capillaries of glomerulus of each bowman's capsule, water, urea, nitrogenous compounds, glucose etc are filtered into the capsule. This process of filtering materials from the glomerulus into the bowman's capsule is called ultra filtration
Answer: filtration
Explanation:
this process allows water and small molecules to filter out of the plasma of the blood.
Explain how the tendinous intersections of the rectus abdominis of the cat differ from those found in humans?
Answer:
The tendinous intersections refer to the three fibrous bands that cross the rectus abdominis muscle. The upper band is located at the xiphoid process, the lower one at the umbilicus level, and the middle one is situated between the two bands. The tendinous intersections in humans are formed of tendinous fibers, which differentiates the rectus abdominis into four components.
On the other hand, the rectus abdominis muscle is spontaneous in the case of cats, that is, it arises from the pubis and merges with the costal cartilages numbering five to seven and the sternum. The alternate overlapping of the internal oblique and rectus abdominis muscles produces the intersections, however, no true intersections are witnessed in it.
14. Valuable molecules like glucose, amino acids, and vitamins are reabsorbed into the blood at which location in the nephron?
A. Loop of Henle
B. Bowman's capsule
C. proximal convoluted tubule
D. collecting duct
E. glomerulus
Answer: Option C) proximal convoluted tubule
Explanation:
The nephron has the proximal tubule where some water, amino acids and salts which are useful to the body are reabsorbed into the blood capillaries against concentration gradient by active transport.
Thus, the process of selective reabsorbtion takes place in the proximal convoluted tubule
Answer: Option C.
Proximal convoluted tubule.
Explanation:
Renal reabsorption is the process molecules like glucose, amino acids, vitamins are reabsorbed from urine preventing them from being loss from the body) (urine)into the circulating blood . Once they are in the convoluted tubule, the molecules diffuse into the blood capillaries. This take place in the nephron.
Blood flow to the skin ________. A) is controlled mainly by decreasing pH B) increases when environmental temperature rises C) increases when body temperature drops so that the skin does not freeze D) is not an important source of nutrients and oxygen for skin cells
Answer:
Blood flow to the skin increases when environmental temperature rises
Explanation:
The blood flow to the skin adapts to both internal and external temperatures, helping the body with thermoregulation while also delivering crucial nutrients and oxygen to skin cells.
Explanation:The blood flow to the skin mainly fluctuates due to variations in body and environmental temperatures. When the environmental temperature rises, the body responds by increasing blood flow to the skin known as vasodilation. This process permits the body to release excess heat. On the other end of the scale, when the body temperature drops, blood flow to the skin decreases, preventing the skin from freezing, a process known as vasoconstriction. Besides its role in thermal regulation, blood flow to the skin is an important source of nutrients and oxygen for skin cells.
Learn more about Blood flow to the skin here:https://brainly.com/question/34311564
#SPJ6
A man of unknown genotype has type B blood, his wife has type A blood (also unknown genotype). List ALL the blood types possible for their children. (you may need to do multiple crosses to consider the different possible genotypes of the parents)
Answer:
AB, aB, Ab
Explanation:
Man has blood type B so could be BB (homozygous dominant) or could be Bb (heterozygous)
Woman has blood type A, so she could be AA (heterozygous dominant) or Aa (heterozygous)
Now, we need to make a separate cross for woman with AA, man with BB and we get children of all AB
Then we can make another cross for genotypes Aa and BB, and we get AB and aB
We can continue this for AA,Bb and for Aa,Bb
Final answer:
Children of a man with type B blood and a woman with type A blood can potentially have A, B, AB, or O blood types, depending on the specific genotypes of their parents. The woman, with type A blood, cannot donate blood to her husband with type B blood, unless her genotype includes the universal donor O allele, which is not specified.
Explanation:
Possible Blood Types for Offspring
The possible blood types for the children of a man with type B blood and a woman with type A blood depend on the genotypes of the parents. Since both parents have blood types that can be either homozygous (AA or BB) or heterozygous (AO or BO), we must consider four different crosses.
If both parents are heterozygous (father is BO and mother is AO), their children could have blood types A, B, AB, or O.
If the father is homozygous (BB) and the mother is heterozygous (AO), their children could have blood types B or AB.
If the father is heterozygous (BO) and the mother is homozygous (AA), their children could have blood types A or AB.
If both parents are homozygous (father is BB and mother is AA), all children would have blood type AB.
Without knowing the exact genotypes of both parents, all we can say is that the children could potentially have any of the ABO blood types.
Nucleotide hydrolysis dehydration monosaccharide polypeptide amino acid polysaccharide polynucleotide
1. A reaction builds polymers from monomers.
2. A reaction breaks down polymers into monomers.
3. A(n) protein hemoglobin.
4. A(n) is a building block of polypeptides, such as the is a polymer of amino acids.
5. A(n) is a building block of polysaccharides such as starch.
6. A(n) is a polymer of monosaccharides. is a building block of polynucleotides such as
7. A(n) is a building block of polynucleotides such as DNA
8. A(n) is a polymer of nucleotides.
A reaction builds polymers from monomers.
Dehydration synthesis reactions build molecules up and require energy, while hydrolysis reactions break molecules down and release energy.
A reaction breaks down polymers into monomers is known as Hydrolysis
Learn more:
https://brainly.com/question/6000641
Consider a pendulum swinging. Which type(s) of energy is/are associated with the pendulum in the following instances: i. the moment at which it completes one cycle, just before it begins to fall back towards the other end, ii. The moment that it is in the middle between the two ends, and iii, just before it reaches the end of one cycle (just before instant i.).
a. i. potential and kinetic, ii. potential and kinetic, iii. Kinetic
b. i. potential, ii. potential and kinetic, iii. potential and kinetic
c. i. potential, ii. kinetic, iii. potential and kinetic
d. i. potential and kinetic, ii. kinetic iii. kinetic
Answer: C. i. potential, ii. kinetic, iii. potential and kinetic
Explanation: Potential energy is the energy that is possessed or exhibited by an object or a body that is static or stable. THE FORMULA FOR POTENTIAL ENERGY IS P.E = MASS OF AN OBJECT* HEIGHT * FORCE OF GRAVITY
Kinetic energy is the energy of an object or a body that is in motion or it is moving from one point to another.
THE FORMULAR FOR KINETIC ENERGY IS 1/2*MASS OF AN OBJECT *VELOCITY OF THE OBJECT SQUARE.
The P and Q genes of a plant are 20 m.u. apart. A PP qq and a pp QQ individual cross, and the F1 cross with each other to produce the F2. The P and p, and Q and q alleles are completely dominant and recessive, and there are no interactions between genes P and Q. What phenotypic ratio would you expect in the F2
Answer:
66 P_ Q_: 9 P_ qq: 9 pp Q_: 16 pp qq
Explanation:
As the genes are 20 m.u apart, recombination frequency between P & Q genes will be 20% (As 1% recombination = 1 m.u).
Parental cross \rightarrow PP QQ x pp qq
Parental gametes \rightarrow PQ, pq
F1 \rightarrow Pp Qq
F1 gametes \rightarrow PQ (Frequency = 0.4), pq (Frequency = 0.4), Pq (Frequency = 0.1), pQ (Frequency = 0.1) (As non-recombinant gametes frequency will be 0.8 & recombinant gametes frequency will be 0.2)
See the attached table.
The phenotypic ratio in the F2 generation would be 3 P Q : 3 p Q or 1:1.
Explanation:In the F2 generation, when individuals from the F1 generation cross with each other, the phenotypic ratio can be determined by using a Punnett square. In this case, the cross is between a PP qq individual and a pp QQ individual.
The gametes produced by the PP qq individual are P q and p Q. The gametes produced by the pp QQ individual are p Q and P q.
When these gametes combine in the F2 generation, the possible genotypes and phenotypes will be as follows:
PP QQ: Phenotype P QPP Qq: Phenotype P QPp QQ: Phenotype P QPp Qq: Phenotype P Qpp QQ: Phenotype p Qpp Qq: Phenotype p QTherefore, the phenotypic ratio in the F2 generation would be 3 P Q : 3 p Q or 1:1.
Learn more about Phenotypic ratio here:https://brainly.com/question/31827445
#SPJ3
When Elodea cells are placed in a hypertonic solution, they lose water and their plasma membranes pull away from the cell wall. What name do we use to describe this phenomenon
Final answer:
The phenomenon observed when Elodea cells are placed in a hypertonic solution, leading to the loss of water and the plasma membrane detaching from the cell wall, is known as plasmolysis.
Explanation:
When Elodea cells are placed in a hypertonic solution, the phenomenon observed is called plasmolysis. This describes the process where the cell loses water and the plasma membrane detaches from the cell wall, resulting in the constriction of the cell membrane.
Unlike animal cells, plant cells have a rigid cell wall that prevents them from bursting in a hypotonic environment, but in a hypertonic environment, the fluid movement is out of the cell causing it to become flaccid as it loses turgor pressure. Plasmolysis can be observed in plant cells that have been underwatered and become wilted due to the absence of adequate turgor pressure to keep the cells turgid and the plant upright.
Two characteristics of natural wines are that they have a maximum alcohol content of 14% and are ""sparkling"" wines. Apply your understanding of alcoholic fermentation to explain these characteristics.
Explanation:
Two attributes of regular wines are that they have a most maximum alcohol content of 14% and are "sparkling" wines.In natural wines, after the ethyl alcohol and are "shimmering" wines. In regular wines, after the ethyl liquor fixation arrives at 14%, the liquor murders the yeast cells, making aging stop. The bubble or shimmer in wines is the collection of CO2. Yeast keep on aging sugars until they either come up short on nourishment or the liquor content gets sufficiently high to slaughter them. The most wine yeasts, 14% is higher than they can tolerate. There are, be that as it may, yeasts that the merchants guarantee can tolerate higher alcoholic rates.Natural wines achieve a maximum alcohol content of around 14% due to the toxicity of ethanol to yeast beyond this concentration. Their sparkling nature is due to the carbon dioxide produced during fermentation, which remains in the wine until opened. Higher alcohol concentrations in beverages like spirits are attained through distillation, beyond what is possible with fermentation alone.
Explanation:The production of natural wines is closely tied to the process of alcoholic fermentation. This natural process, which involves yeast converting sugars into alcohol and carbon dioxide, has limitations, especially in the context of alcohol concentration. A key characteristic of natural wines is that they typically have a maximum alcohol content of around 14%. This limitation arises because the common yeast species used in winemaking, Saccharomyces cerevisiae, cannot survive in environments with ethanol concentrations above 12-14%. Consequently, as the accumulation of ethanol in the wine reaches this level, the yeast dies, preventing further fermentation and higher alcohol content.
Regarding their "sparkling" quality, natural sparkling wines are created by trapping carbon dioxide within the wine. In certain methods, such as in the production of champagne, this is achieved by letting the yeast ferment inside a sealed bottle, generating carbon dioxide that dissolves into the wine under pressure. Upon opening the bottle, the pressure is released, causing the beverage to bubble, creating the sparkling effect. This effervescence is due to the carbon dioxide produced during fermentation, which remains dissolved in the liquid until released.
To produce beverages with higher alcohol content than what natural fermentation yields, a process called distillation is used. Distillation involves heating the fermented solution to evaporate the alcohol, which is then condensed back into liquid form, resulting in a higher concentration of alcohol than achievable by fermentation alone. Spirits such as gin, vodka, and rum are produced through this process, typically containing about 40% alcohol by volume.
Epidemiologic studies that examine the role of a suspected factor in the etiology of a disease may be observational or experimental. The most important difference between experimental and observational studies is that in experimental studies: ___________
a. The study and control groups are equal in size.
b. The study is prospective.
c. The study and control groups are always comparable with respect to all factors other than the exposure.
d. The investigator determines who shall be exposed to the suspected factor and who shall not.
e. Controls are used.
Answer: the correct option is D (The investigator determines who shall be exposed to the suspected factor and who shall not.)
Explanation:
Epidemiology is the branch of medicine that deals with the study of diseases in a defined population to obtain evidence that will aid in the prevention, treatment and control. It is divided into 2 categories:
- experimental studies and
- Observational studies.
In Observational studies, the epidemiologist collects data of interest based on what is seen or observed from the exposure or disease status of each study participants. While in experimental studies, the investigator determines through a controlled process the exposure for each individual and then track over time to detect the effects of the exposure. Therefore in experimental studies,the investigator determines who shall be exposed to the suspected factor and who shall not. I hope this helps, thanks!
A P. bursaria cell that has lost its zoochlorellae is said to be "aposymbiotic." It might be able to replenish its contingent of zoochlorellae by ingesting them without subsequently digesting them. Which of these situations would be most favorable to the re-establishment of resident zoochlorellae, assuming compatible Chlorella are present in P. bursaria's habitat?
A. abundant light, no bacterial prey
B. abundant light, abundant bacterial prey
C. no light, no bacterial prey
D. no light, abundant bacterial prey
The correct answer is B) abundant light, abundant bacterial prey.
The situations that would be most favorable to the re-establishment of resident zoochlorellae, assuming compatible Chlorella are present in P. bursaria's habitat is "abundant light, abundant bacterial prey."
This element needs these conditions of abundant light and bacterial prey to have the correct metabolism and live. The bacterial prey will help the element to grow. This is going to increase the preservation of the green algae.
A research group discovers a new version of happyase, which they call happyase*, that catalyzes the chemical reaction HAPPY SAD. The researchers begin to characterize the enzyme In the first experiment, with [E_t] at 4 nM, they find that the V_max is 1.6 mu M s^-1. In another experiment, with [E]total at 1 nM and [HAPPY] at 30 µM, they find that v0 is 300 nM•s–1. Based on this second experiment, what is the Km for happyase?
Answer:
a. kcat = Vmax/[Et] = (1.6 \muM/s) / 0.004 \muM = 400 s-1
b. Vmax = [Et] kcat = [1 nM] (400 s-1) = 400 nM/s which is 0.4 \muM/s.
Now we can use the Michaelis-Menten equation if all the units are similar (all molar conc in nM or in this case \muM):
V0 = Vmax*[S] / Km+[S] = 0.3 \muM/s = (0.4 \muM/s) (30 \muM) / (Km+30\muM)
Then solving for KM, we get:
(0.3 \muM/s) (Km + 30 \muM) = 0.4\muM/s (30 μM)
0.3 \muM/s(Km) + 9 \muM2/sec = 12 \muM2/s
0.3 \muM/s(Km) = 3 \muM2/s
Km = 10 \muM
Another way to do this would be to first rearrange the Michaelis-Menten equation to:
V0/Vmax = [S] / Km+[S]
(300 nM/s) / (400 nM/s) = 3/4 = [S] / (Km + [S])
4 [S] = 3 Km + 3[S]
Km = [S]/3 = 30 \muM / 3 = 10 \muM
c. After removal of ANGER, the Vmax increased to 4.8 \muM/s and the Km became 15 \muM. It is a mixed because it is affecting both Vmax and Km.
Because Vmax increased by a factor of 3, \alpha'=3.
Similarly, Km varies as a function of \alphaKm/\alpha'. Given that Km increased by a factor of 1.5 when ANGER was removed (that is, the inhibitor decreased the observed Km by 2/3 and \alpha'=3, then \alpha=2.
d. Because both \alpha and \alpha' are affected, ANGER is a mixed inhibitor.
Explanation:
Action potentials are generated at the axon hillock of a neuron and travel in one direction towards the synaptic terminals. Why are action potentials only transmitted in one direction under normal conditions
Answer:
The sodium channel have refractory period.
Explanation:
Action potential are generated at axon hillock of neuron and travel across the neuron in response to stimuli. For the the generation of action potential the steps are stimulus, depolarization, repolarization, hyperpolarization and resting state.
So the first step for action potential after stimulus is depolarization. Depolarization requires the opening of sodium channels to make the inside of neuronal membrane positive. These sodium channels have refractory period. They cannot open immediately after activation. So the sodium channels of one direction in neuron (upstream) become refractory and it is convenent to open the sodium channels of other direction (downstream) of neuron. That why action potential is transmitted in one direction.
It is like the gunpowder placed at a line the fire cannot go back in burnt gun powder direction.
A biology student accidentally loses the labels of two prepared slides she is studying. One is a slide of an intestine, the other of an esophagus. You volunteer to help her sort them out. How would you decide which slide is which?
Answer:
esophagus has stratified squamous
Explanation:
Stratified squamous is a tissue that is found covering and lining parts of the body such as the esophagus. It is known to be many layers of flattened cells.
It performs some functions like the provision of protection from abrasion, pathogens, and chemicals.
It can be found in some parts of the body like surface of skin, linings of mouth, esophagus, rectum,and vagina.
In this case, the presence of stratified squamous in a slide will definitely show that the content in the slide is esophagus. And the slides can then be easily separated and labelled accordingly.
To determine the identity of the slides, you can look for specific features and structures unique to each organ such as villi and goblet cells in the intestine, stratified squamous epithelium, and skeletal muscle fibers in the esophagus.
Explanation:To determine which slide is the intestine and which is the esophagus, you can look for specific features on each slide. The intestine slide may show the presence of villi, which are small finger-like projections that increase the surface area for nutrient absorption. The esophagus slide, on the other hand, may show the presence of stratified squamous epithelium, which is a type of tissue that can withstand mechanical stress.
Another way to differentiate the slides is to look for specific structures associated with each organ. For example, the intestine may have the presence of goblet cells, which secrete mucus, while the esophagus may have the presence of skeletal muscle fibers to aid in peristalsis.
By carefully examining the slides under a microscope and comparing them to known images or descriptions of the structures, you can determine which slide corresponds to the intestine and which corresponds to the esophagus.
Learn more about Identifying slides of the intestine and esophagus here:https://brainly.com/question/32149817
#SPJ2
Animal cells have a membrane that separates the interior of the cell from the outside environment. Typically, an electric potential difference exists between the inner and outer surfaces of the membrane. Consider one such cell where the magnitude of the potential difference is 66 mV, and the outer surface of the membrane is at a higher potential than the inner surface. A sodium ion (Na+) is initially just inside the cell membrane (initially at rest). How much work (in J) is required for a cell to eject the ion, so that it moves from the interior of the cell to the exterior?
The work required to move a sodium ion from inside a cell to the exterior is 1.056 * 10^-21 Joules, using the formula for work where Work = Charge * Potential Difference.
Explanation:This is a question of physics, particularly involving concepts of electricity and work. In order to find the work done to move an ion from the inside of a cell to the exterior, we use the physics formula Work = Charge * Potential Difference.
The charge of a sodium ion (Na+) is equal to the elementary charge, which is approximately 1.6 * 10^-19 Coulombs. The potential difference provided in this problem is 66 mV (millivolts), but we should render this value into volts for the calculation, which would be 0.066 V (volts).
Now, multiplying these values give us:
Work = (1.6 * 10^-19 C) * (0.066 V) = 1.056 * 10^-21 Joules.
Learn more about Work Done in Cellular Processes here:https://brainly.com/question/33230453
#SPJ3
Are most of the cells in our body differentiated or undifferentiated?
Answer: Most cells in the body differentiate.
Explanation:
Cells differentiation is the ability of cells to change form and develop into specialized cells to perform specific functions. These cells in the body retain their ability to divide and form different cells types. Examples include somatic cells such as muscles and skin cells,stem cells and germ line cells which differentiate into gamete cells. Some cells do not differentiate in the body.
For most folks, their knowledge of the importance of microbes is limited to just those microbes that cause sickness. However, you know better! Which of the following describes a true critical feature of microbe biology illustrating their importance? (there can be more than one answer)a.) microbes are very small, and make up a very small portion of the Earth's biomass.b.) they influence our neurobiology and are critical components of ruminant animal digestive systems.c.) some produce oxygen.d.) they are critical for the process of decay.e.) some can fix atmospheric nitrogen which is important for plant growth.f.) they are used in the food and beverage production industry they are important food source for aquatic life
Answer:
Option (b), (c), (d), (e) and (f).
Explanation:
Microbes are the small living organisms that cannot be visible with the naked eye. Microbes plays an important role on earth and has positive as well as negative effect on the other living organisms.
Some microbes can act as pathogen as well. They cover most part of the earth and can be used in food to enhance the texture and taste. The methanogens are present in the gut of ruminant animals and are also useful in the research. Some microorganism has the ability to decompose the biodegradable waste and can even produce oxygen.
Thus, the answer is option (b), (c), (d), (e) and (f).
The importance of the microbe biology critical features statements involved
b.) they influence our neurobiology and are critical components of ruminant animal digestive systems.
c.) some produce oxygen.
d.) they are critical for the process of decay.
e.) some can fix atmospheric nitrogen which is important for plant growth.
f.) they are used in the food and beverage production industry they are an important food source for aquatic life
What is a microbe?Microbes refer to small living organisms that is invisible with the eye. It played a vital role on earth and has positive as well as negative effects on other living organisms. It should act as a pathogen as well. The methanogens should be present in the gut of ruminant animals and are also useful in the research. Some microorganisms should contain the capability to decompose the biodegradable waste and can even generate oxygen.
Learn ore about microbe here: https://brainly.com/question/16329998
A letter to the editor of the Austin American-Statesman, published on December 23, 2009, asks this question: "The trillion-dollar question that Copenhagen has not answered [is this]: Because carbon dioxide molecules arc all identical, why is it that carbon dioxide from carbonated beverages, pets, cattle, farm animals, and humans, yeast, dry ice, fireplaces, charcoal grills, campfires, wildfires, alcohol and ethanol is good, and carbon dioxide from fossil fuel is bad? Can anyone in the United States answer this question?" What is your answer? Your aunt asks you how we know that volcanoes arc not responsible for the observed increase in carbon dioxide. What do you tell her?
Answer:
It's very simple. We all know that the property of a compound never change no matter what the source of that compound is. In case of carbon dioxide it's also the same. Yet, there is one huge difference in emissions of carbon dioxide. The amount of carbon dioxide emitted from all the other sources other than that of fossil fuels is several hundred times less as compared to that of condensed fossil fuels. That's the reason fossil fuels contribute the most in the overall emissions of carbon dioxide all over the world.
The answer is same from volcanic arcs because no volcano is able to emit carbon dioxide amount that is near compare to that of fossil fuels. Furthermore, fossil fuels are burned regularly throughout the world while volcanoes may erupt just a few times in a century.
Compare the movement of nematodes (vinegar eel) with that of annelid worms (earthworm) relating these different kinds of movement to the arrangement of muscle layers in these animals. Why might peristaltic motion be considered an advancement relative to sinusiodal movement?
Answer:
they both use the muscles along side their body which moves them through the ground
Explanation:
1. Annelid worms have two layers of muscles: a layer of longitudinal muscles and a layer of circular muscles. Nematodes have a single layer of longitudinal muscles that runs the length of their body. 2. Peristaltic movement is more forceful than sinusoidal movement, allowing annelid worms to move more easily through thick or viscous materials.
1. Nematodes move through their longitudinal muscles, causing pressure throughout their bodies, and it flexes rather than flattens, moving forward with the thrust.
Circular muscles are one of annelids' three muscular layers. They are found in the outer layer of the body wall and are in charge of restricting the body. Circular muscles are necessary for locomotion, breathing, and excretion.
2. Peristaltic motion:
Furthermore, peristaltic movement permits annelid worms to travel more precisely. This is due to the ability of the longitudinal and circular muscles to be coordinated in order to produce a smooth, wave-like motion. Sinusoidal movement, on the other hand, is jer-kier.
Learn more about nematodes, here:
https://brainly.com/question/32757285
#SPJ4
The compounds that penetrate the human body best are ... Select one: a. the ones that are insoluble in water but are somewhat soluble in grease. b. the ones that are insoluble in both water and grease. c. the ones that are somewhat soluble in water but not grease. d. the ones that are somewhat soluble in both water and grease.
Answer: Option D.the ones that are somewhat soluble in water and grease
Explanation: ionic solvent (polar solvent) readily dissolve in water while oils dissolve in grease. The most common substance that can dissolve in grease and water is soap.
Soap has both hydrophilic and hydrophobic ends which makes it readily dissolve in water and grease. The hydrophilic ends contains a polar head while the hydrophobic ends contains a non polar hydrocarbon tail
You collect the following data on genotypes for a sunflower population:
AA: 15, AB: 70, BB: 15 (so there is an A and a B allele).
Based on Hardy-Weinberg predictions you expected the following numbers:
AA: 25, AB: 50, BB: 25.
Which of the following is a plausible explanation for the deviation?
a) Balancing selection
b) Negative frequency-dependent selection
c) Inbreeding
d) Genetic drift
Answer:
C. InbreedingExplanation:
Breeding of organisms which are closely related genetically (ancestry).
Inbreeding is useful in the retention of desirable characteristic ( which we want to retain) or the elimination of undesirable one (which we do not want to retain).
Indreeding often results in decreased vigour, size, and fertility of the offspring.
Estimate the reduction in bacteria during the passage of wastewater that initially contains 106 organisms per milliliter through three stabilization ponds that are arranged in series. The volumes of the three ponds are 10,000, 20,000 and 6,000 m3, respectively. The flow rate is 1,000 m3/d. Assume that steady-state conditions apply, that ponds are mixed completely because of wind action, that first-order decay kinetics apply
Answer:
an improvement in efficiency is observed with η = 99.99%
Explanation:
the voltage can be calculated using the formula: the detection time can be calculated using the equation:
t1 = V/Q = 10000/1000 = 10 days
t2 = 20000/1000 = 20 days
t3 = 6000/1000 = 6 days
the exit of the organisms will be:
1rd pond = 10^6/(1+(10*10)) = 90909.1 org/mL
2nd pond = 90909.1/(1+(1*20)) = 4329 org/mL
3rd pond = 4329/(1+(1*6)) = 618.4 org/mL
η = ((Cin-Cout)/Cin)*100 = ((10^6-618.4)/(10^6))*100 = 99.93%
detection time t = 12000/1000 = 12 days
Cout1 = 10^6/(1+(1*12)) = 76923.1
Cout2 = 76923.1/(1+(1*12)) = 5917.2
Cout3 = 5917.2/(1+(1*12)) = 455.2
η = ((10^6-455.2)/(10^6))*100 = 99.99%
an improvement in efficiency is observed
16. Most aquatic animals excrete ammonia, while land animals excrete urea or uric acid. What is the most likely explanation for the difference?
A. land animals cannot afford the energy needed to make ammonia
B. they have different diets.
C. ammonia is very toxic and it takes lots of water to dilute it.
D. fish need to get rid of ammonia, but land animals need it to live
E. land animals can get the extra energy needed to make urea or uric acid.
Answer:
It would be C
Explanation:
The most likely explanation for aquatic animals excreting ammonia, and land animals excreting urea or uric acid, is due to ammonia's high toxicity which requires significant amounts of water to dilute, something that is readily available in aquatic environments but not on land.
Explanation:The most likely explanation for the difference in nitrogen waste products excreted by aquatic animals (ammonia) versus land animals (urea or uric acid) is that ammonia is very toxic and requires large amounts of water to dilute it. Aquatic animals are ammonotelic; they excrete ammonia directly into the water, which is a less energy-intensive process and utilizable in an environment where water is abundant. Terrestrial animals, on the other hand, are either ureotelic or uricotelic. Mammals excrete urea, and birds, reptiles, and some terrestrial invertebrates excrete uric acid, primarily because these compounds are less toxic than ammonia and can be safely transported in the body until they can be eliminated. The conversion of ammonia to urea or uric acid requires more energy, but this trade-off is essential for survival on land where water is not as readily available for dilution purposes.
On your first attempt to run a PCR, you realize you forgot to add one of the two primers. The graduate student you work with suggests you throw away your reaction and start over. Why
Explanation:
During PCR, we use two primers one is forward primer and the other one is reverse primer they match the sequence of one one of the two complementary strands of the target DNA, they flank the target region (that the region which we has to be copied). if we add only one primer it copies only one strand of the DNA in multiple copies. Usually we call this as Asymmetric PCR.
Generally Primers are synthetic short stretch of oligonucleotides that are complementary to the target DNA. They act as a foundation for the amplification process of DNA to form multiple copies
The following statement is false. Select the rewording that makes the statement true.
In gymnosperms, seed formation is the process of transferring pollen from a male cone to a female cone.
A) In gymnosperms, pollination is the process of transferring pollen from a male cone to a female cone.
B) In gymnosperms, seed dispersal is the process of transferring pollen from a male cone to a female cone.
C) In gymnosperms, gravitropism is the process of transferring pollen from a male cone to a female cone.
D) In gymnosperms, fertilization is the process of transferring pollen from a male cone to a female cone.
The correct statement is that in gymnosperms, pollination is the process of transferring pollen from a male cone to a female cone, which is necessary for fertilization to occur leading to seed formation.
The original statement is incorrect and needs to be reworded to accurately describe the process in gymnosperms. The correct reworded statement is A) In gymnosperms, pollination is the process of transferring pollen from a male cone to a female cone. Pollination involves the movement of pollen grains from the male reproductive structure to the female reproductive structure. Once the pollen lands on the female cone, it germinates, forming a pollen tube and eventually fertilizing the egg, leading to seed formation. Fertilization is the actual process of the sperm from the pollen uniting with the egg, occurring after pollination has already taken place.